Ordered set vs any set (minimal element and minimum)
up vote
0
down vote
favorite
right now we're doing minimum/maximum/min/max element of sets and relations in class.
Now there's two questions I can't seem to find any idea on how to solve them.
Let M be an ordered Set (we write the relation as ≤)
(1)
Let M be a finite set and let x ∈ M be the only minimal element. Is x the minimum of M?
(2)
Let M be any set and x ∈ M be the only minimal element. Is x the minimum of M?
Now in class we've had the following:
If M is a partially order.
(i) Every minimum of M is minimal in M
(ii) If there exists a minimum in M, then it is the only minimal element of M. Particularly, that means that the minimum is unique.
Well problem (1) and (ii) looks like they're the same thing but (ii) assumes that there exists a minimum in M which isn't the case in (1) but my general idea is that (1) is a true statement and (2) is wrong because we could use an infinite set or just choose a set which has no minimum but one minimal element.
How exactly can you construct an example for (2) to dismiss the statement and how can you proof statement (1) by only using the definition of minimal element/minimum of a set?
discrete-mathematics
asked Nov 22 at 15:55
D idsea J
205
add a comment |
up vote
0
down vote
favorite
right now we're doing minimum/maximum/min/max element of sets and relations in class.
Now there's two questions I can't seem to find any idea on how to solve them.
Let M be an ordered Set (we write the relation as ≤)
(1)
Let M be a finite set and let x ∈ M be the only minimal element. Is x the minimum of M?
(2)
Let M be any set and x ∈ M be the only minimal element. Is x the minimum of M?
Now in class we've had the following:
If M is a partially order.
(i) Every minimum of M is minimal in M
(ii) If there exists a minimum in M, then it is the only minimal element of M. Particularly, that means that the minimum is unique.
Well problem (1) and (ii) looks like they're the same thing but (ii) assumes that there exists a minimum in M which isn't the case in (1) but my general idea is that (1) is a true statement and (2) is wrong because we could use an infinite set or just choose a set which has no minimum but one minimal element.
How exactly can you construct an example for (2) to dismiss the statement and how can you proof statement (1) by only using the definition of minimal element/minimum of a set?
discrete-mathematics
asked Nov 22 at 15:55
D idsea J
205
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
right now we're doing minimum/maximum/min/max element of sets and relations in class.
Now there's two questions I can't seem to find any idea on how to solve them.
Let M be an ordered Set (we write the relation as ≤)
(1)
Let M be a finite set and let x ∈ M be the only minimal element. Is x the minimum of M?
(2)
Let M be any set and x ∈ M be the only minimal element. Is x the minimum of M?
Now in class we've had the following:
If M is a partially order.
(i) Every minimum of M is minimal in M
(ii) If there exists a minimum in M, then it is the only minimal element of M. Particularly, that means that the minimum is unique.
Well problem (1) and (ii) looks like they're the same thing but (ii) assumes that there exists a minimum in M which isn't the case in (1) but my general idea is that (1) is a true statement and (2) is wrong because we could use an infinite set or just choose a set which has no minimum but one minimal element.
How exactly can you construct an example for (2) to dismiss the statement and how can you proof statement (1) by only using the definition of minimal element/minimum of a set?
discrete-mathematics
asked Nov 22 at 15:55
D idsea J
205
right now we're doing minimum/maximum/min/max element of sets and relations in class.
Now there's two questions I can't seem to find any idea on how to solve them.
Let M be an ordered Set (we write the relation as ≤)
(1)
Let M be a finite set and let x ∈ M be the only minimal element. Is x the minimum of M?
(2)
Let M be any set and x ∈ M be the only minimal element. Is x the minimum of M?
Now in class we've had the following:
If M is a partially order.
(i) Every minimum of M is minimal in M
(ii) If there exists a minimum in M, then it is the only minimal element of M. Particularly, that means that the minimum is unique.
Well problem (1) and (ii) looks like they're the same thing but (ii) assumes that there exists a minimum in M which isn't the case in (1) but my general idea is that (1) is a true statement and (2) is wrong because we could use an infinite set or just choose a set which has no minimum but one minimal element.
How exactly can you construct an example for (2) to dismiss the statement and how can you proof statement (1) by only using the definition of minimal element/minimum of a set?
discrete-mathematics
discrete-mathematics
asked Nov 22 at 15:55
D idsea J
205
asked Nov 22 at 15:55
D idsea J
205
asked Nov 22 at 15:55
D idsea J
205
asked Nov 22 at 15:55
D idsea J
205
asked Nov 22 at 15:55
D idsea J
205
205
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0in M$ such that $xnotleq y_0$. However, also $y_0not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $mathbb Z$. It is obviously the only minimal element in $mathbb Zcup{*}$, but not the minimum.
answered Nov 22 at 18:01
Ennar
14.3k32343
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009296%2fordered-set-vs-any-set-minimal-element-and-minimum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0in M$ such that $xnotleq y_0$. However, also $y_0not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $mathbb Z$. It is obviously the only minimal element in $mathbb Zcup{*}$, but not the minimum.
answered Nov 22 at 18:01
Ennar
14.3k32343
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
add a comment |
up vote
1
down vote
accepted
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0in M$ such that $xnotleq y_0$. However, also $y_0not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $mathbb Z$. It is obviously the only minimal element in $mathbb Zcup{*}$, but not the minimum.
answered Nov 22 at 18:01
Ennar
14.3k32343
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0in M$ such that $xnotleq y_0$. However, also $y_0not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $mathbb Z$. It is obviously the only minimal element in $mathbb Zcup{*}$, but not the minimum.
answered Nov 22 at 18:01
Ennar
14.3k32343
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0in M$ such that $xnotleq y_0$. However, also $y_0not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $mathbb Z$. It is obviously the only minimal element in $mathbb Zcup{*}$, but not the minimum.
answered Nov 22 at 18:01
Ennar
14.3k32343
answered Nov 22 at 18:01
Ennar
14.3k32343
answered Nov 22 at 18:01
Ennar
14.3k32343
answered Nov 22 at 18:01
Ennar
14.3k32343
14.3k32343
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
add a comment |
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
Thanks for the answer especially the second part is insightful. How would you construct $y_2$ though?
– D idsea J
Nov 22 at 19:06
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
@D idsea J, you construct $y_{n+1}$ from $y_n$ the same way you construct $y_1$ from $y_0$.
– Ennar
Nov 22 at 22:57
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009296%2fordered-set-vs-any-set-minimal-element-and-minimum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
